Question

Tell how you would prepare a 500.0mL of 0.50 M ammonium carbonate solution? Include all necessary equipment and amount of chemical (in grams).

Answer 1

96,09 g

Explanation:

We start of by calculating the amount of substance `n`, using the equation:
`c=n*V`

Here we insert the known values and we'll get
`n=(0.50 (mol)/L)/(0.5 L)=1 mol`

Then we calculate the amount needed in grams:
`m=n*M_((NH_4)2CO_3)`

We insert the calculated `n` and the molar mass of ammonium carbonate:
`m=1 mol * 96.09 g/(mol) = 96,09 g`

So the make the solution, you'll need:

• Scale
• volumetric flask 500 ml
• pipette or plastic pipette
• deionised water

Start of by weighing the 96.09 g of ammonium carbonate. Than transfere the ammonium carbonate to a 500 ml volumetric flask. Start adding deionised water, up to the neck of the volumetric flask and add the rest of the water with pipette (could be a plastic pipette) up till the mark on the volumetric flask.
Put on a lid and mix the solution by rotation the flask up and down (180 degrees).

Answer 2

Well, get yourself a `500*mL` volumetric flask as shown....

Explanation:

And then add an accurate mass of approx. `25*g` of ammonium carbonate....that is the mass has to be around `25*g`...you weigh by difference, and you should have a mass to three places after the decimal....say you transferred `24.0010*g`...

And then half fill the flask with distilled water, and swirl, until all of the salt goes into the solution....and then make it to volume to the graduation point on the flask....invert the beast several times...to assure homogeneous mixing.

`(NH_4)_2CO_3=((24.0010*g)/(96.09*g*mol^-1))/(0.500*L)=0.4998*mol*L^-1`

Please note that this salt is not especially stable...